Sunday, April 19, 2015

Physics 9702 Doubts | Help Page 116

  • Physics 9702 Doubts | Help Page 116



Question 588: [Temperature > Internal energy]
(a) On Fig.1, place tick () against those changes where internal energy of the body is increasing.
water freezing at constant temperature           ..........................................
a stone falling under gravity in a vacuum       ..........................................
water evaporating at constant temperature     ..........................................
stretching a wire at constant temperature       ..........................................

(b) A jeweller wishes to harden a sample of pure gold by mixing it with some silver so that mixture contains 5.0% silver by weight. Jeweller melts some pure gold and then adds the correct weight of silver. Initial temperature of the silver is 27 °C. Use data of Fig.2 to calculate initial temperature of the pure gold so that the final mixture is at the melting point of pure gold.
gold                 silver
melting point / K                                                                     1340                1240
specific heat capacity (solid or liquid) / J kg–1K–1                   129                  235
specific latent heat of fusion / kJ kg–1                                     628                  105

(c) Suggest a suitable thermometer for measurement of the initial temperature of the gold in (b).

Reference: Past Exam Paper – June 2003 Paper 4 Q2



Solution 588:
(a)
water freezing at constant temperature           ..........................................
a stone falling under gravity in a vacuum       ..........................................
water evaporating at constant temperature                  tick
stretching a wire at constant temperature                    tick
{The concept of ‘Internal Energy’
Internal energy is the sum of potential and kinetic energy due to the RANDOM motion of the molecules.
It is important to point out that the internal energy is due to the energies of the randomly moving molecules only AND as the word says internal energy is due to the ‘internal’ structure of the system (due to molecules or atoms), not the whole system itself.

For example, for a car moving with speed v or moving up a slope. The sum of its kinetic energy and gain in potential energy do NOT give the internal energy. It only gives the total energy of the car.

So, again when we are talking about internal energy, we are firstly considering the atoms / molecules inside the materials and secondly, we are only considering those atoms / molecules that are in random motion.}

{Case 1: water freezing at constant temperature
For constant temperature, the kinetic energy of the molecules is unchanged. For freezing, there is a change in phase from liquid to solid occurring. This requires energy to form the new bonds. So, the material is actually losing some of its potential energy to form the bonds. So, internal energy is decreasing in this case.

Case 2: a stone falling under gravity in a vacuum
This case is dealing with the energy of a large object, not its internal structure. As explained before, in such a case, we do not talk about internal energy, but rather the total energy of the stone.

Case 3: water evaporating at constant temperature
Again for constant temperature, the kinetic energy of the molecules is unchanged. For evaporation, there is a change in phase from liquid to gas occurring. The potential energy between the randomly moving molecules as they go in the gas state (more molecules are now in random motion). So, internal energy is increasing in this case.

Case 4: stretching a wire at constant temperature
Again for constant temperature, the kinetic energy of the molecules is unchanged. For stretching a wire, work needs to be done on the wire. Since the wire has now become longer than its initial length, the molecules / atoms inside the wire have more space to move freely / randomly compared to its initial shape where all the molecules / atoms were packed together (in the solid wire). We say that the wire has elastic potential energy. So, the internal energy of the wire increases.}

(b)
{To reach the final temperature which is equal to the melting point of pure gold, the temperature of the silver (initially at 300K) should increase. This means that the temperature of the pure gold should have decreased to provide energy to the silver.
ΔQ = mcΔθ
m is the total mass. 95% of m is gold and 5% of m is silver.}
Heat lost by liquid gold = 0.95m × 129 × ΔT
{The melting point go gold is higher than that of silver. So, before reaching the final temperature, the silver has melted first. This requires energy which is given by ΔQ = mL = 0.05m × 105 000 where L is the specific latent heat of fusion.
After melting, the silver needs to increase its temperature to the final temperature. This further requires energy given by ΔQ = mcΔθ = 0.05m × 235 × (1340 – 300)}
Heat gained (by silver) = 0.05m × 235 × (1340 – 300) + 0.05m × 105 000
{Heat lost by liquid gold = heat gained by silver}
122.5mΔT = 17 470m
Change in temperature ΔT = 143 K
{Final temperature is the melting point of gold}
Initial temperature of gold = 143 + 1340 = 1483 K

(c) Example: thermocouple / resistance thermometer










Question 589: [Current of Electricity]
Four resistors of equal value are connected as shown.

How will powers to the resistors change when resistor W is removed?
A The powers to X, Y and Z will all increase.
B The power to X will decrease and the powers to Y and Z will increase.
C The power to X will increase and the powers to Y and Z will decrease.
D The power to X will increase and the powers to Y and Z will remain unaltered.

Reference: Past Exam Paper – June 2011 Paper 11 Q36



Solution 589:
Answer: C.
Power dissipated in a resistor = I2R

Connecting 2 resistors of, say, resistance R in series causes the overall resistance to increase (= R+R = 2R).

Connecting 2 resistors of, say, resistance R in parallel causes the overall resistance to decrease (= [1/R + 1/R]-1 = 0.5R).

Since the 4 resistors have the same resistance in the connection shown, the p.d. across each parallel connection (and each resistor) will be the same.

Overall resistance in circuit = 0.5R + 0.5R = R
Let the e.m.f. in the circuit = E
Current in the circuit = E / R = I

The total current splits at the parallel combination.
Power dissipated in X = (I/2)2R = I2R / 4 = 0.25 I2R
The same power is dissipated in all the resistors.


When W is removed, the overall resistance of the parallel combination of W and X changes from 0.5R to R (due to X only).

Total resistance in new circuit = R + 0.5R = 1.5R
Current in circuit = E / 1.5R = (2/3)I
The overall current in the circuit will also change – it decreases since the overall resistance in the complete circuit has increased. This current flows through X but is split into half at the combination of Y and Z. So, the current through Y and Z is half that through X.

Power dissipated = I2R
Power dissipated in X = [(2/3)I]2R = 4 I2R / 9 = 0.44 I2R

Half the current flows through Y and Z. The same power is dissipated through both resistors.
Power dissipated in Y = [0.5 (2/3)I]2R = I2R / 9 = 0.11 I2R

So, the power in X increases and the power in Y (and Z) decreases.










Question 590: [Electromagnetism > Hall Probe]
(a) Constant current is maintained in a long straight vertical wire. Hall probe is positioned a distance r from the centre of the wire, as shown in Fig.1.


(i) Explain why, when Hall probe is rotated about the horizontal axis XY, the Hall voltage varies between a maximum positive value and a maximum negative value.
(ii) Maximum Hall voltage VH is measured at different distances r.
Data for VH and the corresponding values of r are shown in Fig.2.
VH / V             r / cm
0.290               1.0
0.190               1.5
0.140               2.0
0.097               3.0
0.073               4.0
0.060               5.0

It is thought that VH and r are related by expression of the form
VH = k / r
where k is a constant.
1. Without drawing a graph, use data from Fig.2 to suggest whether the expression is valid.
2. A graph showing variation with 1 / r of VH is plotted.
State features of the graph that suggest that the expression is valid.

(b) Hall probe in (a) is now replaced with a small coil of wire connected to sensitive voltmeter. Coil is arranged so that its plane is normal to the magnetic field of the wire.
(i) State Faraday’s law of electromagnetic induction and hence explain why voltmeter indicates a zero reading.
(ii) State three different ways in which an e.m.f. may be induced in the coil.

Reference: Past Exam Paper – June 2010 Paper 41 Q5



Solution 590:
(a)
(i) The Hall voltage VH depends on the angle between (the plane of) the probe and B-field (magnetic field).
either The Hall voltage VH is maximum when the plane and the B-field are normal to each other.
or The Hall voltage VH is zero when the plane and the B-field are parallel
or The Hall voltage VH depends on the sine of the angle between the plane and B-field

(ii)
1.
{From the formula, the product of VH and r is a constant. k = VH r}
Calculate VH r at least three times {any sets can be chosen}
1st set: 0.29 × 1.0 = 0.29 (to 2sf) = 0.3
2nd set: 0.19 × 1.5 = 0.285 = 0.29 (to 2sf) = 0.3
3rd set: 0.14 × 2.0 = 0.28 (to 2sf) = 0.3
EITHER If given to 1 s.f., the product is constant, so the expression is valid or approximately constant so valid

OR If given to 2 s.f., the product is not constant, so the expression is invalid

2. The graph is a straight line that passes through the origin

(b)
(i) Faraday’s law of electromagnetic induction states that the e.m.f. induced is proportional / equal to the rate of change of (magnetic) flux (linkage).
The voltmeter indicates a zero reading because there is a constant field in coil / flux (linkage) of coil does not change

(ii) Example:
Vary the current (in wire) / switch current on or off / use a.c. current
Rotate the coil
Move the coil towards / away from wire










Question 591: [Work, Energy and Power]
Box of weight 30 N is released from rest on a ramp that is at angle of 30° to the horizontal. The box slides down the ramp so that it falls through a vertical distance of 8.0 m. A constant frictional force of 10 N acts on box while it is moving.

What is the kinetic energy of box after falling through this distance?
A 80 J                          B 160 J                        C 240 J                        D 400 J

Reference: Past Exam Paper – November 2014 Paper 13 Q17



Solution 591:
Answer: A.
In this question, there is a frictional force of 10N that opposes the motion of the box down the ramp. So, work needs to be done against friction.

Consider a case where the ramp was frictionless. From the conservation of energy, the gravitational potential energy would be converted into kinetic energy of the box as it falls.

But in this case, friction is present. So, not all of the gravitational potential energy would be kinetic energy. Some of the energy is used up as work done against friction.

Loss in GPE = Gain in KE + Work against friction

Loss in GPE = mgh = (mg)h = (30) (8) = 240 J

Let the distance travelled by the box along the ramp = s
sin30 = 8 / s
Distance s = 8 / sin30 = 16m

Work done against friction = Fs = 10 (16) = 160 J

Loss in GPE = Gain in KE + Work against friction
240 = Gain in KE + 160
Gain in KE = 240 – 160 = 80 J



15 comments:

  1. Dear Sir, please consider answering these questions.

    01/O/N/07 Q.40

    02/O/N/08 Q.7(b)(ii)

    21/O/N/09 Q.6(b)

    12/M/J/10 Q.6

    21/M/J/10 Q.3(b)(iii)part1

    22/M/J/10 Q.4(b)(ii),(c), Q.5(a)(ii), Q.7(b)(ii)2.

    23/M/J/10 Q.7(b)

    11/O/N/10 Q.1,24,31

    12/O/N/10 Q.26,33

    21/O/N/10 Q.3(c)(ii),Q.7(b)(ii)2.

    22/O/N/10 Q.4(b)(ii)

    11/M/J/11 Q.9,10,14,16,23,24,25,27,32,34

    21/M/J/11 Q.7(b)

    Your cooperation is highly appreciated.

    ReplyDelete
    Replies
    1. For 01/O/N/07 Q.40, check question 598 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-118.html

      Delete
    2. Due to the large number of questions in the post above, please consider answering (in details) only paper 2 questions as these are the first priority. You may answer the rest of the questions after one month from now. Thanks

      Delete
    3. For 02/O/N/08 Q.7(b)(ii), check at
      http://physics-ref.blogspot.com/2014/09/9702-november-2008-paper-2-worked.html

      Delete
    4. Sir, please consider answering the rest of the paper 2 by now.

      Delete
    5. For 21/O/N/09 Q.6(b), check solution 612 at
      http://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-121.html

      Delete
    6. This comment has been removed by the author.

      Delete
    7. For 22/M/J/10 Q.4(b)(ii),(c), check at
      http://physics-ref.blogspot.com/2014/10/9702-june-2010-paper-22-worked.html

      Delete
  2. Hello..can you help me with questions from oct/nov 2009 paper 12 questions 8,9,11,12, 25 and 27
    Thank you

    ReplyDelete
    Replies
    1. Check if they are explained at
      http://physics-ref.blogspot.com/2014/11/9702-november-2009-paper-12-worked.html

      if not, ask the specific questions there

      Delete
    2. thank you..i checked and found all explanation but not Q9
      kindly help me with Q9

      Delete
  3. Dear Sir, please consider answering these questions.

    02/O/N/08 Q.7(b)(ii)

    21/O/N/09 Q.6(b)

    21/M/J/10 Q.3(b)(iii)part1

    22/M/J/10 Q.4(b)(ii),(c), Q.5(a)(ii), Q.7(b)(ii)2.

    23/M/J/10 Q.7(b)

    21/O/N/10 Q.3(c)(ii),Q.7(b)(ii)2.

    22/O/N/10 Q.4(b)(ii)

    21/M/J/11 Q.7(b)

    ReplyDelete
    Replies
    1. See the links in the comments above

      Delete
  4. Here are the remaining questions for now:
    21/M/J/10 Q.3(b)(iii) part 1

    22/M/J/10 Q.5(a)(ii), Q.7(b)(ii)2.

    23/M/J/10 Q.7(b)

    21/O/N/10 Q.3(c)(ii),Q.7(b)(ii)2.

    22/O/N/10 Q.4(b)(ii)

    ReplyDelete
    Replies
    1. For 21/M/J/10 Q.3(b)(iii) part 1
      http://physics-ref.blogspot.com/2014/10/9702-june-2010-paper-21-worked.html

      Delete

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 116